Problem: The lifespans of sloths in a particular zoo are normally distributed. The average sloth lives $17.8$ years; the standard deviation is $2.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a sloth living less than $22.6$ years.
Solution: $17.8$ $15.4$ $20.2$ $13$ $22.6$ $10.6$ $25$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $17.8$ years. We know the standard deviation is $2.4$ years, so one standard deviation below the mean is $15.4$ years and one standard deviation above the mean is $20.2$ years. Two standard deviations below the mean is $13$ years and two standard deviations above the mean is $22.6$ years. Three standard deviations below the mean is $10.6$ years and three standard deviations above the mean is $25$ years. We are interested in the probability of a sloth living less than $22.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the sloths will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the sloths will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $13$ years and the other half $({2.5\%})$ will live longer than $22.6$ years. The probability of a particular sloth living less than $22.6$ years is ${95\%} + {2.5\%}$, or $97.5\%$.